UVA 11235 Frequent values（RMQ）

原题链接

Problem Description

You are given a sequence of n integers a1, a2, … , an in non-decreasing order. In addition to that, you
are given several queries consisting of indices i and j (1 ≤ i ≤ j ≤ n). For each query, determine the
most frequent value among the integers ai
, … , aj .

Input

The input consists of several test cases. Each test case starts with a line containing two integers n and
q (1 ≤ n, q ≤ 100000). The next line contains n integers a1, … , an (−100000 ≤ ai ≤ 100000, for each
i ∈ {1, …, n}) separated by spaces. You can assume that for each i ∈ {1, … , n − 1}: ai ≤ ai+1. The
following q lines contain one query each, consisting of two integers i and j (1 ≤ i ≤ j ≤ n), which
indicate the boundary indices for the query.
The last test case is followed by a line containing a single ‘0’.

Output

For each query, print one line with one integer: The number of occurrences of the most frequent value
within the given range.
Note: A naive algorithm may not run in time!

Sample Input

10 3
-1 -1 1 1 1 1 3 10 10 10
2 3
1 10
5 10
0

Sample Output

1
4
3

题目大意

给出一个非降序数列，对于每个询问(i,j)，回答ai,ai+1，ai+2···aj中出现次数最多的数字的个数。

解题思路

使用Sparse-Table算法，d[i][j]表示从第i位开始，长度为2j的区间中的最大值，编码时要考虑边界情况。

AC代码

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<vector>#include<string>#include<queue>#include<list>#include<stack>#include<set>#include<map>#define ll long long#define ull unsigned long long#define rep(i,a,b) for (int i=(a),_ed=(b);i<_ed;i++)#define fil(a,b) memset((a),(b),sizeof(a))#define cl(a) fil(a,0)#define pb push_back#define PI 3.1415927#define inf 0x3f3f3f3fusing namespace std;int a[100005];int num[100005];int lefta[100005];int righta[100005];int d[100005][32];vector<int> counta;int main(void){    int n,q;    while(cin>>n&&n)    {        cl(num);        cl(lefta);        cl(righta);        counta.clear();        cin>>q;        int pos=0;        int st=0;        int ed=n-1;        int jishu=1;        scanf("%d",&a[0]);        num[0]=0;        lefta[0]=0;        if(n==1) counta.pb(1);        rep(i,1,n)        {            scanf("%d",&a[i]);            if(a[i]==a[i-1])            {                num[i]=pos;                lefta[i]=st;                jishu++;            }            else            {                counta.pb(jishu);                jishu=1;                num[i]=++pos;                st=i;                lefta[i]=st;            }        }        if(jishu>=2) counta.pb(jishu);        righta[n-1]=n-1;        ed=n-1;        for(int i=n-2;i>=0;--i)        {            if(a[i]==a[i+1])            {                righta[i]=ed;            }            else            {                ed=i;                righta[i]=ed;            }        }        int ac=counta.size();        rep(i,0,ac) d[i][0]=counta[i];        for(int j=1;(1<<j)<=ac;++j)        {            for(int i=0;i+(1<<j)-1<ac;++i)            {                d[i][j]=max(d[i][j-1],d[i+(1<<(j-1))][j-1]);            }        }        rep(i,0,q)        {            scanf("%d%d",&st,&ed);            st--;            ed--;            int L,R,res;            L=num[st]+1;            R=num[ed]-1;            if(L-2==R) printf("%d\n",ed-st+1);            else if(L-1==R) printf("%d\n",max(righta[st]-st+1,ed-lefta[ed]+1));            else            {                res=-inf;                int k=0;                int ans;                while(1<<(k+1)<=R-L+1) k++;                ans=max(d[L][k],d[R-(1<<k)+1][k]);                res=max(res,ans);                res=max(res,righta[st]-st+1);                res=max(res,ed-lefta[ed]+1);                printf("%d\n",res);            }        }    }    return 0;}